Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long int  
using namespace std;

int main(void)
{ 

	ll n,i,toss;
	double heads,s;
	cin>>n;
	vector <double> dp(n+1,0);
	dp[0]=1;
	for(toss=0;toss<n;toss++)
	{
		cin>>heads;
		for(i=toss+1;i>=0;i--)
		{
			if(i==0)
			s=0;
			else s=dp[i-1]*heads;
			
			dp[i]=s+dp[i]*((double)(1-heads));
			
		}
	}
	double ans=0;
	for(i=(n+1)/2;i<=n;i++)
	ans+=dp[i];
	
	cout << fixed << setprecision(10)<<ans;

}